Q. In an experiment on photoelectric emission from a metallic surface, wavelength of incident light is $2 \times 10^{- 7}$m and stopping potential is 2.5 V. The threshold frequency of the metal (in Hz ) approximately (charge on electron $e = 1.6 \times 10^{- 19}$ C, Planck’s constant $h = 6.6 \times 10^{- 34}$J-s)
A) $12 \times 10^{15}$
B) $9 \times 10^{15}$
C) ${9 \times 10}^{14}$
D) $12 \times 10^{13}\ \ $
Explanation
(c) $eV_{0} = hv - hv_{0}$
$\therefore$ Threshold frequency,
$v_{0} = v - \frac{eV_{0}}{h}$
= $\frac{c}{\lambda} - \frac{eV_{0}}{h}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v_{0} = \frac{3 \times 10^{8}}{2 \times 10^{- 7}} - \frac{1.6 \times 10^{- 19} \times 2.5}{6.6 \times 10^{- 34}}$
$= \ 9.0 \times 10^{14}\ Hz$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.